\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 214, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2013 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2013/214\hfil Ground state solutions] {Ground state solutions for semilinear problems with a Sobolev-Hardy term} \author[X. Chen, W. Chen \hfil EJDE-2013/214\hfilneg] {Xiaoli Chen, Weiyang Chen} % in alphabetical order \address{Xiaoli Chen \newline Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022, China} \email{littleli\_chen@163.com} \address{Weiyang Chen (corresponding author)\newline Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022, China} \email{xiaowei19901207@126.com} \thanks{Submitted April 19, 2013. Published September 26, 2013.} \subjclass[2000]{35J60, 35J65} \keywords{Existence; ground state; critical Hardy-Sobolev exponent; \hfill\break\indent semilinear Dirichlet problem} \begin{abstract} In this article, we study the existence of solutions to the problem \begin{gather*} -\Delta u= \lambda u+\frac{|u|^{2_s^\ast-2}u}{|y|^s}, \quad x\in \Omega,\\ u = 0, \quad x\in \partial \Omega, \end{gather*} where $\Omega$ is a smooth bounded domain in $\mathbb{R}^N$ $(N\geq3)$. We show that there is a ground state solution provided that $N=4$ and $\lambda_m<\lambda<\lambda_{m+1}$, or that $N\geq5$ and $\lambda_m\leq\lambda<\lambda_{m+1}$, where $\lambda_m$ is the m'th eigenvalue of $-\Delta$ with Dirichlet boundary conditions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \allowdisplaybreaks \section{Introduction} Let $\Omega$ be a smooth bounded domain of $\mathbb{R}^N=\mathbb{R}^k\times\mathbb{R}^{N-k}$, where $2\leq k0 \quad \text{in } \mathbb{R}^N,\quad u\in D^{1,2}(\mathbb{R}^N) up to a constant. If$s=0$, Equation \eqref{eq:1.2} becomes the Sobolev inequality, for which best constant was computed, and proved existence of minimizers in \cite{Aubin} and \cite{Talenti}. In the case$s=2$, \eqref{eq:1.2} still holds true, it is an extension of the Hardy inequality. In the more general case$0\leq s<2$with$k=N$, the best constant was obtained in \cite{GGMT}, and minimizers were found in \cite{Lieb}, which are radially symmetric. Therefore, it can be shown by using ODEs, see \cite{Lieb}, that up to dilations and translations, minimizers take the form $\frac 1{(1+|x|^{2-s})^{\frac {N-2}{2-s}}}.$ It is noted that equation \eqref{eq:1.3} is invariant with respect to the scalings and$z$-translations; that is,$u$is a solution of \eqref{eq:1.3} if only if$u_\alpha(x) = \alpha^{(N-2)/2}u(\alpha y,\alpha(z-z_0))$,$\alpha>0$, satisfies the equation. Hence, problem \eqref{eq:1.3} has lack of the compactness. In the case$0< s <2$,$2\leq k0$, and$S$is achieved by the concentration-compactness principle. So problem \eqref{eq:1.3} has a positive solution in$D^{1,2}(\mathbb{R}^N)$. Since the minimizer of problem \eqref{eq:1.2} can not be radially symmetric, they cannot be found among solutions of ODEs, but of PDEs. This brings difficulties to find exact forms of the minimizer. In the particular case$s=1$, problem \eqref{eq:1.3} becomes $$\label{eq:1.4} -\Delta u=\frac{u^{\frac{N}{N-2}}}{|y|},\quad u>0 \quad \text{in } \mathbb{R}^N,\quad u\in D^{1,2}(\mathbb{R}^N).$$ By the moving plane method, it was proved in \cite{FMS} that all solutions of \eqref{eq:1.4} are cylindrically symmetric. Thus, problem \eqref{eq:1.4} can be reduced to an elliptic equation in the positive cone in$\mathbb{R}^2$, and it was shown in \cite{FMS} that$u$is a solution of \eqref{eq:1.4} if and only if $$\label{eq:1.5} u(y,z)=\lambda^{(N-2)/2}V(\lambda y,\lambda(z+z_0))$$ for some$\lambda>0$and$z_0\in\mathbb{R}^{N-k}$, where $$\label{eq:1.6} V(x)=V(y,z)=\frac{C_{N,k}}{\left((1+|y|)^2+|z|^2\right)^{(N-2)/2}} =\frac{\left((N-2)(k-1)\right)^{(N-2)/2}}{\left((1+|y|)^2 +|z|^2\right)^{(N-2)/2}}.$$ This result allows one to obtain existence results for problem \eqref{eq:1.1} in the case$s = 1$. Denote by$0<\lambda_1,\dots,\lambda_k,\dots$the eigenvalues of$-\Delta$with zero Dirichlet boundary condition. When$0<\lambda<\lambda_1$and$s = 1$, it was proved in \cite{AWZ} and \cite{FabbriPHD} that there exists a solution of problem \eqref{eq:1.1} by the mountain pass lemma and constrained variation respectively. In this article, we consider the existence of solutions to problem \eqref{eq:1.1} for general case$0C_1>0$such that $$\label{eq:1.7} \frac{C_1}{1+|x|^{N-2}}\leq u(x)\leq\frac{C_2}{1+|x|^{N-2}}.$$ This estimate suffices to serve our purpose. Using the Nehari manifold method introduced in \cite{P}, and developed in \cite{SW}, we show the following result. \begin{theorem}\label{thm:1.0} Let$ N=4$and$\lambda_m<\lambda<\lambda_{m+1}$or$N\geq5$and$\lambda_m\leq\lambda<\lambda_{m+1}$for some$m\in\mathbb{N}$, then there exists a ground state solution of problem \eqref{eq:1.1}. \end{theorem} In section 2, we describe a variational framework to study the ground state solution of problem \eqref{eq:1.1}. We prove Theorem \ref{thm:1.0} in section 3. \section {Preliminaries} Denote by$E=H^1_0(\Omega)$the Hilbert space with the scalar product $\langle u,v\rangle=\int_\Omega\nabla u\nabla v\,dx$ and the induced norm$\|\cdot\|$. Let$(\varphi_j,\lambda_j)$be the eigenfunctions and eigenvalues of$-\Delta$in$\Omega$with zero Dirichlet boundary condition. Suppose that$m$is a fixed positive integer and$\lambda_m\leq\lambda<\lambda_{m+1}$, we define the subspaces$E^-=\operatorname{span}\{\varphi_1,\dots,\varphi_m\}$and$E^+=\operatorname{span}\{\varphi_j,j\geq m\}$of$E$, then$E=E^+\oplus E^-$. The functional associated to problem \eqref{eq:1.1} is defined by $J(u)=\frac12\int_\Omega|\nabla u|^2\,dx -\frac\lambda2\int_\Omega|u|^2\,dx -\frac1{2_s^\ast}\int_\Omega\frac{|u|^{2_s^\ast}}{|y|^s}\,dx$ for$u\in H^1_0(\Omega)$, which is$C^1$and critical points of$J$are solutions of problem \eqref{eq:1.1}. To find ground state solutions of \eqref{eq:1.1}, we introduce as \cite{P} a submanifold of$E$. Define $$\label{eq:2.1} \mathcal{N}=\{u\in E\setminus{\{0\}}:\langle {\nabla J(u)},u\rangle=0, {\nabla J(u)}\in E^+ \}.$$ The set$\mathcal{N}$is the intersection of the standard Nehari manifold$\{u\in E\setminus{\{0\}}:\langle {\nabla J(u)},u\rangle=0\}$with the pre-image$(\nabla J)^{-1}(E^+)$. \begin{proposition}\label{prop:2.1} The set$\mathcal{N}$is a$C^1$submanifold of$E$with codimension$m+1$. Moreover, every critical point of the restriction$J|_\mathcal{N}$is a nontrivial critical point of the functional$J$. \end{proposition} \begin{proof} The result can be proved as \cite{SWW}, see also \cite{S}. We sketch the proof here for reader's convenience. Let$F:E\setminus{\{0\}}\to \mathbb{R}\times E^-$be a map defined by $F(u)=(\langle {\nabla J(u)},u\rangle,Q\nabla J(u)),$ where$Q$is the orthogonal projection of$E$onto$E^-$, then$\mathcal{N}=F^{-1}(0)$. Consider the inner product $(t_1,z_1)\cdot(t_2,z_2)=t_1t_2+\langle z_1,z_2\rangle \quad \text{for } t_1,t_2\in \mathbb{R},\; z_1,z_2\in E^-.$ We claim that for every$(t,z)\in \mathbb{R}\times E^-$,$(t,z)\neq(0,0)$, the inequality $$\label{eq:2.2} (DF(u)(tu+z))\cdot(t,z)<0$$ holds. This implies the first part of the proposition. Now, we prove the claim. Indeed, for$(t,z)\neq(0,0)$, since $$\langle {\nabla J(u)},u\rangle=\langle {\nabla J(u)},z\rangle=0,$$ we deduce that $$\label{eq:2.3} \begin{split} &(DF(u)(tu+z))\cdot(t,z)\\ &=\Big(\int_{\Omega}|\nabla z|^2\, dx-{\lambda}\int_{\Omega}|z|^2 \,dx\Big)\\ &\quad-\int_{\Omega}\Big((2_s^\ast-2)t^2|u|^2+2(2_s^\ast-2)tzu +(2_s^\ast-1)|z|^2\Big)\frac{|u|^{2_s^\ast-2}}{|y|^s}dx. \end{split}$$ For$\lambda_m\leq\lambda<\lambda_{m+1}$, it is readily verified that \eqref{eq:2.2} holds. Next, we verify as in \cite{SWW} that$w\in E$is a critical point of$J$if and only if$u\in \mathcal{N}$and$DJ(u)|_{T_u\mathcal{N}}=0$. The proof is complete. \end{proof} We recall that a ground state solution$u$to \eqref{eq:1.1} is any element of$\mathcal{N}$such that$DJ(u)$vanishes on$T_u\mathcal{N}$and$J(u)=c$, where $$\label{eq:2.4} c=\inf_{\mathcal{N}}J.$$ By the argument in \cite{SW}, for every$v\in E^+\setminus{\{0\}}$, there is a unique continuous map pair$(f(v),g(v))\in (0,\infty)\times E^-$such that$F(f(v)v+g(v))=0$and $J(f(v)v+g(v))=\max_{t>0,z\in E^-}J(tv+z).$ Hence, $$\label{eq:2.5} c=\inf_{\mathcal{N}}J=\inf_{v\neq0,v\in E^+}J(f(v)v+g(v)) =\inf_{\{v\neq0,v\in E^+\}}\max_{\{t>0,z\in E^-\}}J(tv+z).$$ \section{Existence results} In this section, we show that problem \eqref{eq:2.4} is achieved. The minimizer of problem \eqref{eq:2.4} is actually a ground state solution of \eqref{eq:1.1}. Let $$\label{eq:3.1} S=\inf_{u\in E,u\not=0}\bigg\{\frac{\int_{\Omega}|\nabla u|^2\,dx}{(\int_{\Omega}\frac{u^{2_s^\ast}}{|y|^s}\,dx)^{2/2_s^\ast}}\bigg\}.$$ We know from \cite{BT} that$S$can be achieved, which is independent of$\Omega$and depends only by$N,k,s$, moreover the infimum$S$is never achieved when$\Omega$is a bounded domain, we denote the minimizer by$U(x)>0$. By \eqref{eq:1.7}, $$\frac{C_1}{1+|x|^{N-2}}\leq U(x)\leq\frac{C_2}{1+|x|^{N-2}}.$$ The following elementary lemma is readily verified. \begin{lemma}\label{lem:3.1} Suppose$A>0$,$B>0$. Then $\max_{t>0}(A\frac{t^2}2-B\frac{t^{2_s^\ast}}{2_s^\ast}) =\frac{2-s}{2(N-s)}\Big(\frac{A}{B^{2/2_s^\ast}}\Big)^{\frac{N-s}{2-s}}.$ \end{lemma} \begin{lemma}\label{lem:3.2} Suppose that $$\label{eq:3.2} c<\frac{2-s}{2(N-s)}S^{\frac{N-s}{2-s}},$$ then there exists$v\in E^+\setminus{\{0\}}$such that $\max_{t>0,w\in E^-}J(tv+w)=J(f(v)v+g(v))=c.$ \end{lemma} \begin{proof} Take any sequence$\{v_n\}$in$E^+\setminus{\{0\}}$such that$\|v_n\|=1$and $$\label{eq:3.3} \max_{t>0,w\in E^-}J(tv_n+w)\to c.$$ Without loss of generality, we can assume that \begin{gather*} v_n\rightharpoonup v \quad \text{in } E^+,\\ v_n \to v \quad \text{in } L^{2}(\Omega), \\ v_n\to v \quad\text{a.e. }\Omega. \end{gather*} Suppose $A=\lim_{n\to \infty}\int_{\Omega}|\nabla (v_n-v)|^2 dx\,,\quad B=\lim_{n\to \infty}\int_{\Omega}\frac{|v_n-v|^{2_s^\ast}}{|y|^s}dx\,.$ Using the Brezis-Lieb's Lemma, from \eqref{eq:3.3} we obtain $$\label{eq:3.4} J(tv+w)+\frac 1 2 At^2-\frac 1{2_s^\ast}B t^{2_s^\ast}\leq c, \quad \forall t>0, \; \forall w\in E^-.$$ If$v=0$and$B=0$, from the assumption$\|v_n\|=1$, we deduce that$A=1$. Hence$t^2\leq2c-2J(w)$for every$t>0$and every$w\in E^-$, a contradiction. Assume now$B\neq0$. From Lemma \ref{lem:3.1}, we obtain that $$\label{eq:3.5} \frac{2-s}{2(N-s)}S^{\frac{N-s}{2-s}}\leq\frac{2-s}{2(N-s)} \Big(\frac A {B^{ 2 /{2_s^\ast}}}\Big)^{\frac{N-s}{2-s}} =\max_{t>0}\Big(\frac 1 2 At^2-\frac 1{2_s^\ast}B t^{2_s^\ast}\Big).$$ If$v=0$, we obtain from \eqref{eq:3.2}, \eqref{eq:3.4} and \eqref{eq:3.5} that $\frac{2-s}{2(N-s)}S^{\frac{N-s}{2-s}} \leq c<\frac{2-s}{2(N-s)}S^{\frac{N-s}{2-s}},$ a contradiction. Thus$v\neq0$. Denote$h=g(v)/f(v)$. It follows from the definition of$c$that $$\label{eq:3.6} \begin{split} c&\leq J(f(v)(v+h))=\max_{t>0}J(t(v+h))\\ &=\frac{2-s}{2(N-s)}\bigg\{\frac {\int_{\Omega}|\nabla (v+h)|^2\,dx -{\lambda}\int_{\Omega}|v+h|^2dx}{ \big(\int_{\Omega} \frac{|v+h|^{2_s^\ast}}{|y|^s}dx\big)^{2/2_s^\ast}} \bigg\}^{\frac{N-s}{2-s}}. \end{split}$$ By \eqref{eq:3.4} and Lemma \ref{lem:3.1}, $$\label{eq:3.7} \begin{split} c&\geq \max_{t>0}\Big(J(t(v+h))+\frac 1 2 At^2-\frac 1{2_s^\ast}B t^{2_s^\ast} \Big)\\ &=\frac{2-s}{2(N-s)}\bigg\{\frac {A+\int_{\Omega}|\nabla (v+h)|^2\,dx -{\lambda}\int_{\Omega}|v+h|^2\,dx} { \big(B +\int_{\Omega}\frac{|v+h|^{2_s^\ast}}{|y|^s}\,dx\big)^{2/2_s^\ast}} \bigg\}^{\frac{N-s}{2-s}}. \end{split}$$ Putting together \eqref{eq:3.2}, \eqref{eq:3.5}, \eqref{eq:3.6} and \eqref{eq:3.7}, we obtain \label{eq:3.8} \begin{split} &\Big(\frac{2(N-s)}{2-s}c\Big)^{\frac{2-s}{N-s}} \Big(B +\int_{\Omega}\frac{|v+h|^{2_s^\ast}}{|y|^s}\,dx\Big)^{2/2_s^\ast}\\ &<\Big(\frac{2(N-s)}{2-s}c\Big)^{\frac{2-s}{N-s}} \Big(B^{2/2_s^\ast} +\Big(\int_{\Omega}\frac{|v+h|^{2_s^\ast}}{|y|^s}\,dx \Big)^{2/2_s^\ast}\Big)\\ &0$, we define $U_\varepsilon(x)=\varepsilon^{\frac{2-N}2}U(\frac{x-(0,z_0)}{\varepsilon})$, $u_\varepsilon=\varphi(x)U_\varepsilon(x)$, Then $u_\varepsilon\in E$ for $\varepsilon>0$ small. We have following estimates for $u_{\varepsilon}$. \begin{lemma}\label{lem:3.3} Suppose $N\geq3$, we have \begin{gather}\label{eq:3.9} \|u_{\varepsilon}\|^2=\|U\|^2+O(\varepsilon^{N-2})+O(\varepsilon^{N-s}),\\ \label{eq:3.10} \int_{\Omega}\frac{|u_{\varepsilon}|^{2_s^\ast}}{|y|^{s}}dx =\int_{R^N}\frac{|U|^{2_s^\ast}}{|y|^{s}}dx+O(\varepsilon^{N-s}),\\ \label{eq:3.11} \int_{\Omega}|u_{\varepsilon}(x)|^{2}dx \geq \begin{cases} C\varepsilon^{2}+O(\varepsilon^{N-2}),& N\geq5,\\ C\varepsilon^{2}|\ln\varepsilon|+O(\varepsilon^2),& N=4,\\ C\varepsilon+O(\varepsilon^2),& N=3, \end{cases} \\ \label{eq:3.12} \int_{\Omega}u_{\varepsilon}(x)dx\leq C\varepsilon^{(N-2)/2},\\ \label{eq:3.13} \int_{\Omega}\frac{|u_{\varepsilon}|^{2_s^\ast-1}}{|y|^{s}}dx \leq C\varepsilon^{(N-2)/2}. \end{gather} \end{lemma} \begin{proof} First, we estimate \eqref{eq:3.10}. There holds \begin{align*} \int_{\Omega}\frac{|u_{\varepsilon}|^{2_s^\ast}}{|y|^{s}}dx &=\int_\Omega\frac{|\varphi U_{\varepsilon}|^{2_s^\ast}}{|y|^{s}}dx =\int_{\Omega}\frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx-\int_{\Omega}(1-\varphi^{2_s^\ast})\frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx\\ &=\int_{R^{N}}\frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx-\int_{R^{N}\backslash\Omega}\frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx- \int_{\Omega}(1-\varphi^{2_s^\ast})\frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx\\ &=\int_{R^{N}}\frac{U^{2_s^\ast}}{|y|^{s}}dx-\int_{R^{N}\backslash\Omega}\frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx- \int_{\Omega\setminus B_{\frac\rho2}(0,z_0)}(1-\varphi^{2_s^\ast})\frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx.\\ \end{align*} Since $\int_{R^{N}\backslash B_{R}(0,z_{0})} \frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx \leq\int_{R^{N}\backslash\Omega}\frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx \leq\int_{R^{N}\backslash B_{\rho}(0,z_{0})} \frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx,$ while \begin{align*} \int_{R^{N}\backslash B_{R}(0,z_{0})}\frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx &=\int_{R^{N}\backslash B_{R}(0,z_{0})}\varepsilon^{s-N} \frac{U(\frac{x-(0,z_{0})}{\varepsilon})^{2_s^\ast}}{|y|^{s}}dx\\ &=\int_{R^{N}\backslash B_{R}(0)}\varepsilon^{s-N} \frac{U(\frac{x}{\varepsilon})^{2_s^\ast}}{|y|^{s}}dx\\ &\le C\varepsilon^{s-N}\int_{R^{N}\backslash B_{R}(0)} \Big(\frac{1}{1+|\frac x{\varepsilon}|^{N-2}}\Big)^{2_s^\ast}\frac1{|y|^{s}}dx\\ &= C\varepsilon^{N-s}\int_{R^{N}\backslash B_{R}(0)} \Big(\frac{1}{\varepsilon^{N-2}+|x|^{N-2}}\Big)^{2_s^\ast}\frac1{|y|^{s}}dx\\ &=O(\varepsilon^{N-s}), \end{align*} and similarly, $\int_{R^{N}\backslash B_{\rho}(0,z_{0})}\frac{U_{\varepsilon}^{2_s^\ast}}{|y|^{s}}dx=O(\varepsilon^{N-s}).$ Thus, we obtain $\int_{\Omega}\frac{|u_{\varepsilon}|^{2_s^\ast}}{|y|^{s}}dx=\int_{R^{N}}\frac{U^{2_s^\ast}}{|y|^{s}}dx+O(\varepsilon^{N-s}).\\$ That is, \eqref{eq:3.10} holds. Next, we estimate \eqref{eq:3.11}. In fact, \begin{align*} \int_{\Omega}|u_{\varepsilon}|^{2}dx &=\int_{\Omega}\varphi^{2}|U_{\varepsilon}|^{2}dx \leq\int_{B_{\rho}(0,z_{0})}{|U_{\varepsilon}|}^{2}\,dx\\ &=\varepsilon^{2-N}\int_{B_{\rho}(0)}{U(\frac{x}{\varepsilon})}^{2}dx\\ &\leq\varepsilon^{2-N}\int_{B_{\rho}(0)} \frac{C}{{\left(1+|\frac{x}{\varepsilon}|^{N-2}\right)}^{2}}dx\\ &=\varepsilon^{N-2}\int_{B_{\rho}(0)}\frac{C}{{\left(\varepsilon^{N-2}+|x|^{N-2}\right)}^{2}}dx\\ &\leq\varepsilon^{N-2}\int_{B_{\varepsilon}(0)}\frac{C}{\varepsilon^{2(N-2)}}dx+\varepsilon^{N-2}\int_{B_{\rho}(0)\backslash B_{\varepsilon}(0)}\frac{C}{|x|^{2(N-2)}}dx\\ &=\begin{cases} C\varepsilon^{2}+O(\varepsilon^{N-2}),& N\geq5,\\ C\varepsilon^{2}|\ln\varepsilon|+O(\varepsilon^2),& N=4,\\ C\varepsilon+O(\varepsilon^2),& N=3. \end{cases} \end{align*} Now, we estimate \eqref{eq:3.9}. Observe that $\int_{\Omega}|\nabla u_{\varepsilon}|^{2}\,dx =\int_{\Omega}U_{\varepsilon}^{2}|\nabla\varphi|^{2}dx +\int_{\Omega}\nabla U_{\varepsilon}\nabla(\varphi^{2}U_{\varepsilon})\,dx$ and $-\Delta U_{\varepsilon}= U_{\varepsilon}^{2\ast-1}/|y|^{s}$, we find $\int_{\Omega}\nabla U_{\varepsilon}\nabla(\varphi^{2}U_{\varepsilon})\,dx =\int_{\Omega}\varphi^{2}\frac{U_{\varepsilon}^{2\ast}}{|y|^{s}}\,dx$ and $\int_{\Omega}|\nabla u_{\varepsilon}|^{2}\,dx =\int_{\Omega}|\nabla\varphi|^{2}U_{\varepsilon}^{2}\,dx +\int_{\Omega}\varphi^{2}\frac{U_{\varepsilon}^{2\ast}}{|y|^{s}}.$ Since $\nabla\varphi=0$ in $B_\rho(0,z_0)$, we have \begin{align*} \int_{\Omega}|\nabla\varphi|^{2}U_{\varepsilon}^{2}\,dx &=\int_{\Omega\setminus B_\rho(0,z_0)}|\nabla\varphi|^{2}U_{\varepsilon}^{2}\,dx\\ &\leq\int_{B_R(0,z_0)\setminus B_\rho(0,z_0)}|\nabla\varphi|^{2} U_{\varepsilon}^{2}\,dx\\ &\leq C\int_{B_R(0,z_0)\setminus B_\rho(0,z_0)}U_{\varepsilon}^{2}\,dx\\ &\leq C\int_{B_R(0)\setminus B_\rho(0)}\varepsilon ^{2-N} \frac 1{(1+|\frac{x}\varepsilon|^{N-2})^2}\,dx=O(\varepsilon^{N-2}). \end{align*} On the other hand, we can show that $\int_{\Omega}\varphi^{2}\frac{|U_{\varepsilon}|^{2_s^\ast}}{|y|^{s}}dx =\int_{R^{N}}\frac{U^{2_s^\ast}}{|y|^{s}}dx+O(\varepsilon^{N-s}) =\int_{R^N}|\nabla U|^2\,dx+O(\varepsilon^{N-s}).$ Therefore, $\int_{\Omega}|\nabla u_\varepsilon|^2\,dx =\|\nabla U\|^2+O(\varepsilon^{N-2})+O(\varepsilon^{N-s}).$ Now, we estimate \eqref{eq:3.12}. \begin{align*} \int_{\Omega}u_{\varepsilon}dx &=\int_{B_{\rho}(0,z_0)}\varepsilon^{\frac{2-N}{2}} U(\frac{x-(0,z_0)}{\varepsilon})\\ &\leq C\int_{B_{\rho}(0)}\varepsilon^{\frac{2-N}{2}}\frac{1}{1+| \frac{x}{\varepsilon}|^{N-2}}dx\\ &=\varepsilon^{\frac{N-2}{2}}\int_{B_{\rho}(0)} \frac{1}{\varepsilon^{N-2}+|x|^{N-2}}dx\\ &\leq C\varepsilon^{\frac{N-2}{2}}\int_{B_{\varepsilon}(0)} \frac{1}{\varepsilon^{N-2}}dx+ C\varepsilon^{\frac{N-2}{2}}\int_{B_{\rho}(0)\backslash{B_{\varepsilon}(0)}} \frac{1}{|x|^{N-2}}dx\\ &\leq C\varepsilon^{\frac{N-2}{2}}. \end{align*} Finally, there holds \begin{align*} \int_{\Omega}\frac{|u_{\varepsilon}|^{2_s^\ast-1}}{|y|^{s}}\,dx &\leq \int_{B_\rho(0,z_0)}\frac{|U_{\varepsilon}|^{2_s^\ast-1}}{|y|^{s}}\,dx\\ &\leq \varepsilon^{\frac{N+2-2s}2}\int_{B_\rho(0)} \Big(\frac{1}{\varepsilon^{N-2}+|x|^{N-2}}\Big)^{2_s^\ast-1}\frac{dx}{|y|^{s}}\\ &=\varepsilon^{\frac{N+2-2s}2}\int_{B_\varepsilon(0)} \Big(\frac{1}{\varepsilon^{N-2}+|x|^{N-2}}\Big)^{2_s^\ast-1} \frac{dx}{|y|^{s}}\\ &\quad+\varepsilon^{\frac{N+2-2s}2}\int_{{B_\rho(0)}\setminus{B_\varepsilon(0)}} \Big(\frac{1}{\varepsilon^{N-2}+|x|^{N-2}}\Big)^{2_s^\ast-1}\frac{dx}{|y|^{s}}\\ &\leq C\varepsilon^{(N-2)/2} +\varepsilon^{\frac{N+2-2s}2}\int_{{B_\rho(0)} \setminus{B_\varepsilon(0)}}\frac{1}{(|y|^2+|z|^2)^{\frac{N+2-2s}2}} \frac{dx}{|y|^{s}} \end{align*} and \begin{align*} &\varepsilon^{\frac{N+2-2s}2}\int_{{B_\rho(0)}\setminus{B_\varepsilon(0)}} \frac{1}{(|y|^2+|z|^2)^{\frac{N+2-2s}2}}\frac{dx}{|y|^{s}}\\ &=\varepsilon^{\frac{N+2-2s}2}\int_{({B_\rho(0)} \setminus{B_\varepsilon(0)})\cap\{x=(y,z):|y|\geq|z|\}} \frac{1}{(|y|^2+|z|^2)^{\frac{N+2-2s}2}}\frac{dx}{|y|^{s}}\\ &\quad+\varepsilon^{\frac{N+2-2s}2}\int_{({B_\rho(0)}\setminus{B_\varepsilon(0)})\cap\{x=(y,z):|y|<|z|\}}\frac{1}{(|y|^2+|z|^2)^{\frac{N+2-2s}2}}\frac{dx}{|y|^{s}}\\ &\leq C\varepsilon^{\frac{N+2-2s}2}\int_{\{x=(y,z):\frac{\varepsilon}{\sqrt 2}<|y|,|z|<\rho\}}\frac{1}{|z|^{N+2-2s}}\frac{dx}{|y|^{s}}\\ &\quad+C\varepsilon^{\frac{N+2-2s}2}\int_{\{x=(y,z):\frac{\varepsilon}{\sqrt 2}<|y|,|z|<\rho\}}\frac{1}{|y|^{N+2-2s}}\frac{dx}{|y|^{s}}\\ &\leq C\varepsilon^{(N-2)/2}, \end{align*} which implies \eqref{eq:3.13}. \end{proof} \begin{proposition}\label{prop:3.1} There holds $c<\frac{2-s}{2(N-s)}S^{\frac{N-s}{2-s}}.$ \end{proposition} \begin{proof} We will check that $$\label{eq:3.14} \max_{t>0,v\in E^-}J(t u_\varepsilon+v)<\frac{2-s}{2(N-s)}S^{\frac{N-s}{2-s}}.$$ Let $\omega=\Omega\setminus{\rm supp}\varphi$. By \cite[Lemma 3.3]{SWW}, $v\mapsto\|v\|_{L^{2_s^\ast}(\omega)}$ defines a norm on $E^-$. Since $\dim E^-=m<+\infty$, all the norms are equivalent on $E^-$. For every $t>0$ and every $v\in E^-$, by convexity we deduce \label{eq:3.15} \begin{aligned} &\int_{\Omega}\frac{|tu_{\varepsilon}(x)+v(x)|^{2_s^\ast}}{|y|^s}\, dx\\ &=\int_{\Omega\setminus\omega}\frac{|tu_{\varepsilon}(x)+v(x)|^{2_s^\ast}}{|y|^s} \,dx+\int_{\omega}\frac{|v(x)|^{2_s^\ast}}{|y|^s}\, dx\\ &\geq t^{2_s^\ast}\int_{\Omega}\frac{|u_{\varepsilon}(x)|^{2_s^\ast}}{|y|^s}\, dx+{2_s^\ast}t^{2_s^\ast-1}\int_{\Omega}\frac{|u_{\varepsilon}(x)|^{2_s^\ast-1}v(x)}{|y|^s}\, dx +{2_s^\ast}C_1\|v\|^{2_s^\ast}. \end{aligned} It follows that \label{eq:3.16} \begin{aligned} J(t u_\varepsilon+v) &\leq J(t u_\varepsilon)+t\int_{\Omega} \nabla u_\varepsilon\nabla v+{\frac 12}\int_{\Omega}|\nabla v|^2\,dx\\ &\quad -\lambda t\int_\Omega u_{\varepsilon}(x)v(x)\,dx -{\frac {\lambda} 2}\int_{\Omega}|v(x)|^2\,dx\\ &\quad-t^{2_s^\ast-1}\int_{\Omega}\frac{|u_{\varepsilon}(x)|^{2_s^\ast-1}v(x)}{|y|^s} dx-C_1\|v\|^{2_s^\ast}.\\ \end{aligned} By the assumption $\lambda_m\leq\lambda<\lambda_{m+1}$, $$\label{eq:3.17} \int_{\Omega}|\nabla v|^2\,dx-{\lambda} \int_{\Omega}|v(x)|^2\,dx\leq(\lambda_m-\lambda)\|v\|^2\leq0.$$ In particular, we can write $J(t u_\varepsilon+z)\leq A(t^2+t\|v\|+t^{2_s^\ast-1}\|v\|) -B(t^{2_s^\ast}+\|v\|^{2_s^\ast})$ for suitable constants $A>0$ and $B>0$. Hence there exists $R>0$ such that, for $\varepsilon$ small, $t>R$ and $v\in E^-$ there holds $J(t u_\varepsilon+v)\leq0$. On the other hand, whenever $t\leq R$, $$\label{eq:3.18} J(t u_\varepsilon+v)\leq J(t u_\varepsilon) +O\big(\varepsilon^{(N-2)/2}\big)\|v\| -C_1\|v\|^{2_s^\ast}\leq J(t u_\varepsilon) +O\Big(\varepsilon^{\frac {(N-2)(N-s)}{N+2-2s}}\Big).$$ Indeed, integrating by parts and using the definition of $E^-$, we obtain $$\label{eq:3.19} \begin{split} &\int_{\Omega}\nabla u_\varepsilon\nabla v\,dx -{\lambda}\int_{\Omega}u_{\varepsilon}(x) v(x)\,dx\\ &=\int_\Omega(-\Delta v)u_\varepsilon\,dx -{\lambda}\int_{\Omega}u_{\varepsilon}(x) v(x)\,dx\\ &\leq|\lambda_m-\lambda|\int_{\Omega}|u_\varepsilon(x)v(x)|\,dx \leq|\lambda_m-\lambda|\|v(\cdot)\|_{L^\infty}\int_{\Omega}|u_\varepsilon(x)|\,dx\\ &\leq C|\lambda_m-\lambda|\|v\|\int_{\Omega}|u_{\varepsilon}(x)|\,dx \end{split}$$ and \begin{eqnarray*} \big|\int_{\Omega}\frac{|u_{\varepsilon}(x)|^{2_s^\ast-1}v}{|y|^s}\,dx\big| \leq C\|v\|\int_{\Omega}\frac{|u_{\varepsilon}(x)|^{2_s^\ast-1}}{|y|^s}\,dx. \end{eqnarray*} By \eqref{eq:3.12} and \eqref{eq:3.13} , we get $\int_{\Omega}|u_{\varepsilon}(x)|\,dx \leq C\varepsilon^{(N-2)/2}, \quad \int_{\Omega}\frac{|u_{\varepsilon}(x)|^{2_s^\ast-1}}{|y|^s}\,dx \leq C\varepsilon^{(N-2)/2} .$ By the Young inequality, $O\Big(\varepsilon^{(N-2)/2}\Big)\|v\| \leq O\Big(\varepsilon^{\frac {(N-2)(N-s)}{N+2-2s}}\Big)+C_1\|v\|^{2_s^\ast}.$ Therefore, together with \eqref{eq:3.17}, we see that \eqref{eq:3.18} holds. Since $N\geq 5$ implies $\frac {(N-2)(N-s)}{N+2-2s}>2$. By Lemma \ref{lem:3.2}, for $\varepsilon>0$ small enough, \begin{align*} &\max_{t>0, v\in E^-}J(t u_\varepsilon+v)\\ &\leq \max_{t>0}J(t u_\varepsilon) +O\Big(\varepsilon^{\frac {(N-2)(N-s)}{N+2-2s}}\Big)\\ &=\frac {2-s}{2(N-s)}\bigg(\frac {\| u_\varepsilon\|^2 -\lambda \|u_{\varepsilon}(x)\|^2_{L^2(\Omega)}} {(\int_{\Omega}\frac{|u_{\varepsilon}(x)|^{2_s^\ast}}{|y|^s}\,dx)^{2/2_s^\ast}} \bigg)^{\frac{N-s}{2-s}} +O\Big(\varepsilon^{\frac {(N-2)(N-s)}{N+2-2s}}\Big)\\ &\leq\frac {2-s}{2(N-s)}\Big(S-C\lambda\varepsilon^{2} +O(\varepsilon ^{N-2}) \Big)^{\frac{N-s}{2-s}} +O\Big(\varepsilon^{\frac {(N-2)(N-s)}{N+2-2s}}\Big)\\ &<\frac {2-s}{2(N-s)}S^{\frac{N-s}{2-s}}. \end{align*} Assume now that $N=4$. From \eqref{eq:3.12} and \eqref{eq:3.13} , we obtain $\int_{\Omega}|u_{\varepsilon}(x)|\,dx\leq C\varepsilon, \quad \int_{\Omega}\frac{|u_{\varepsilon}(x)|^{2_s^\ast-1}}{|y|^s}\,dx \leq C\varepsilon.$ By the assumption $\lambda_m<\lambda<\lambda_{m+1}$, $$\label{eq:3.20} \int_{\Omega}|\nabla v|^2\,dx-{\lambda} \int_{\Omega}|v(x)|^2\,dx \leq(\lambda_m-\lambda)\|v\|^2=-C_2\|v\|^2.$$ Inequality \eqref{eq:3.16}, \eqref{eq:3.19} and \eqref{eq:3.20} imply that, for $t\leq R$, $J(t u_\varepsilon+v)\leq J(t u_\varepsilon) +O(\varepsilon)\|v\|-C_2\|v\|^{2}\leq J(t u_\varepsilon)+O(\varepsilon^2).$ From Lemma \ref{lem:3.2}, for $\varepsilon>0$ small enough, we obtain \begin{align*} &\max_{t>0, v\in E^-}J(t u_\varepsilon+v)\\ &\leq\frac {2-s}{2(4-s)} \bigg(\frac {\|u_\varepsilon\|^2-\lambda \|u_{\varepsilon}\|^2_{L^2(\Omega)}} {(\int_{\Omega}\frac{|u_{\varepsilon}|^{2_s^\ast}}{|y|^s}\,dx)^{2/2_s^\ast}} \bigg)^{\frac{4-s}{2-s}} +O(\varepsilon^2)\\ &\leq\frac {2-s}{2(4-s)} \bigg(\frac{\|U\|^2+O\left(\varepsilon^2\right) -\lambda\left(C\varepsilon^2\left|\ln\varepsilon\right|+O(\varepsilon^2)\right)} {\big(\int_{R^N}\frac{|U|^{2_s^\ast}}{|y|^s}\,dx +O\left(\varepsilon ^{4-s}\right)\big)^{2/(4-s)} } \bigg)^{\frac {4-s}{2-s}} +O(\varepsilon^2)\\ &\leq\frac {2-s}{2(4-s)} \left(S-C\lambda\varepsilon^2\left|\ln\varepsilon\right|+O\left(\varepsilon ^2\right) \right)^{\frac{4-s}{2-s}}+O(\varepsilon^2)\\ &<\frac {2-s}{2(4-s)}S^{\frac{4-s}{2-s}}. \end{align*} \end{proof} \begin{proof}[Proof of Theorem \ref{thm:1.0}] By Lemma \ref{lem:3.1} and Proposition \ref{prop:3.1}, there exists $u\in \mathcal{N}$ such that $J(u)=c$ and $DJ(u)|_{T_u\mathcal{N}}=0$. 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